Question

# The graph which depicts the result of Rutherford's gold foil experiment with $\alpha -$particle is$\theta$: Scattering angle$Y$: Number of scattered $\alpha -$ particles detected(Plots are schematic and not to scale)

A
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B
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C
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D
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Solution

## The correct option is B Step 1: Given data:$\theta$: Scattering angle$Y$: Number of scattered $\alpha -$ particles detectedStep 2: Formula Used:According to Rutherford's observations from the gold foil experiment-$N\left(\theta \right)=\frac{K}{{\mathrm{sin}}^{4}\frac{\theta }{4}}$Where, $K$ is constant, $N\left(\theta \right)$ is no. of alpha particles scattered through an angle $\theta$Step 3: Computing the graph between $Yand\theta$:From the scattering formula,$N\left(\theta \right)=\frac{K}{{\mathrm{sin}}^{4}\frac{\theta }{4}}$Here, $Y$: number of scattered $\alpha -$ particles detected$Y\alpha \frac{1}{{\left(\mathrm{sin}\frac{\theta }{4}\right)}^{4}}$When, $\theta =2\pi \phantom{\rule{0ex}{0ex}}\frac{\theta }{4}=\frac{\pi }{2}$Since $0to2\pi$ function ${\mathrm{sin}}^{4}\frac{\theta }{4}$increases so $\frac{1}{{\mathrm{sin}}^{4}\frac{\theta }{4}}$decreasesSo, At $\theta \to 0,\mathrm{sin}0=0,1}{{\mathrm{sin}}^{4}\theta /4}=\infty \phantom{\rule{0ex}{0ex}}$Thus the graph will decrease exponentially.Note that this figure which depicts the result of Rutherford's gold foil experiment with $\alpha -$particles will beTherefore the $\alpha -$particle is$Y\alpha \frac{1}{{\left(\mathrm{sin}\frac{\theta }{2}\right)}^{4}}$.Hence,option B is the correct answer.

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